chapter 30: the nature of the atom - physics.umanitoba.ca
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Chapter 30: The Nature of the Atom
• Rutherford scattering and the nuclear atom –!the nucleus is really " small!
• Atomic line spectra – wavelengths characteristic of each element
• Bohr model of the hydrogen atom –!" quantization of angular momentum ! quantization of energy
• x-rays
• The laser
• Omit 30.5, 6 – quantum mechanical picture of H-atom, Pauli
! Exclusion Principle, periodic table
1Monday, April 2, 2007
Early model of the atom
• About 10-10 m in size.
• Positively charged (“pudding”), with negatively charged electrons " (“plums”) embedded in the pudding – “plum pudding model”
Stability of the plum pudding was questionable.
Where would the characteristic spectral lines come from?
#
2Monday, April 2, 2007
Geiger+Marsden: Fall of the plum pudding
Alpha (!) particles – nuclei of 4He atom, emitted by some radioactive nuclei – were scattered from a thin gold foil and observed on a screen as flashes of light.
Far more were scattered at large angle than would be possible with the weak electric field inside a “plum pudding” atom.
Rutherford: realized that the positive charges of the atom had to be contained in a very small volume – the nucleus –!electric field very strong close to it.
" planetary model of the atom with electrons " in orbit around nucleus
3Monday, April 2, 2007
Very schematic picture of an atom
Size of the nucleus $ 10-15 – 10-14 m
Size of the atom $ 10-10 m
Equal amounts of + and – charges in the neutral atom.
Electrons are in orbit around the nucleus in a planetary model of the atom.
(charge +Ze)
4Monday, April 2, 2007
Problem with a Planetary Model of the Atom
• The electrons suffer centripetal acceleration in their orbital motion.
• Accelerated charges should radiate electromagnetic energy.
" The electrons should lose energy and spiral into the nucleus in very " " little time.
" A planetary atom should not be stable!
" Classical theory does not explain the structure of the atom.
" Small systems, such as atoms, must behave differently from
! large.
30.15Monday, April 2, 2007
Fraunhofer absorption lines from the sun
http://www.harmsy.freeuk.com/fraunhofer.html
1 Å (Angstrom) = 0.1 nm
• Fraunhofer lines – due to absorption of "sunlight " by elements in the atmosphere of the sun • chemical elements emit and absorb light at
! the same wavelengths
• Models of the atom need to explain this
The sun: blackbody radiation for T $ 6000 K
6Monday, April 2, 2007
Line spectrum of the hydrogen atom
InfraredUltraviolet Visible
m Series
1 Lyman ultraviolet
2 Balmer visible
3 Paschen infrared
4 Brackett infrared
Balmer found by trial and error a simple formula to calculate the wavelength of all lines of the hydrogen atom:
R = Rydberg constant = 1.097 % 107 m-1
1
!= R
(1
m2− 1
n2
)m= 1,2,3 . . .
n= m+1,m+2,m+3 . . .
n = 4n = 5n = 6
n = & series limit
n = 3
n = 4n = 2
m = 1 m = 2 m = 3
+ other series
7Monday, April 2, 2007
1
!= 1.097×107
(1
12− 1"
)= 1.097×107
! = 91.2 nm (series limit)
Spectrum of H-atom
Example: Lyman series, m = 1:
For n = &:
For n = 2:
1
!= 1.097×107
(1
12− 1
22
)= 0.823×107
! = 121.5 nm
R = Rydberg constant = 1.097 % 107 m-1
1
!= R
(1
m2− 1
n2
)m= 1,2,3 . . .
n= m+1,m+2,m+3 . . .
(series limit, shortest wavelength in Lyman series)
(longest wavelength in Lyman series)
m = 1
n = 2n = &
m–1
m–1
8Monday, April 2, 2007
Bohr model of the hydrogen atom
Now superseded by more modern quantum mechanical ideas, but gives the correct answers.
Assume:• A planetary model with electron " in orbit around the nucleus.• There are certain electron orbits " that are stable (“stationary states”). " This is contrary to classical theory " and is not explained in this model.• Light is emitted or absorbed when an electron changes state.• Energy is conserved, so the energy of the photon is the difference in " energy between the initial and final states:
" E = hf = Ei – Ef " – what are the allowed energies?
E = hf
“Stationary” states
= Ei – Ef
9Monday, April 2, 2007
The electron is kept in orbit by the attractive Coulomb force between it and the nucleus.
F =kZe
2
r2=mv
2
r
mv2 =
kZe2
r
The potential energy is:
PE=−kZe2
r
And so the total mechanical energy, E = KE + PE, is:
But, what are the “good” values of r corresponding to the stationary states?
= 2 KE (A)
E =kZe
2
2r− kZe
2
r=−kZe
2
2r(B)
Energies of hydrogen atom in the Bohr model
A hydrogen-like atom or ion with just one electron in orbit
m
10Monday, April 2, 2007
Combine with (A): mv2r = kZe
2
So, v2 =
kZe2
mr=
[nh
2!mr
]2
r =1
kZe2m
[nh
2!
]2= (5.29×10−11 m)
n2
Z
En =−[2!2mk2e4
h2
]Z2
n2, n= 1,2,3 . . .
Ln = mvr =nh
2!n = 1, 2, 3...
Energies of hydrogen atom in the Bohr model
And then, substituting into (B), the total mechanical energy is:[E =−kZe
2
2r
]
En =−13.6 Z2
n2eV
Assertion (Bohr): the angular momentum of the electron around the nucleus can have only certain, quantized, values:
m
n = “quantum number”
11Monday, April 2, 2007
Bohr Model for H and H-like Atoms
• A planetary model in which the electron defies the normal laws of " electromagnetism and does not radiate when in certain “stationary” " states.
• The angular momentum of the electron in the stationary states is:
" n is a “quantum number”.
• The energy of the n’th quantum state is:
• Light is emitted or absorbed when the atom changes energy state.
• The energy of the photon is the difference in energy of the two states " of the atom.
En = −13.6Z2
n2eV
Ln = mvr =nh
2π, n = 1, 2, 3 . . .
12Monday, April 2, 2007
Emission of a photon
Emission and absorption occur at the same wavelengths (as seen in the Fraunhofer absorption lines of sunlight).
E = hfEi = Ef +h f
Absorption of a photon
E = hf
Ef +h f = Ei
13Monday, April 2, 2007
Spectrum of hydrogen-like atoms and ions
Energy levels: En =−13.6 Z2
n2eV
E = hf
ni
nf
(Just one electron in orbit around the
nucleus)
In agreement with Balmer’s formula[R=
(13.6×1.602×10−19 J)hc
= 1.097×107 m−1]
= Ei – Ef
Energy of photons: E = Ei − Ef =hc
λ= 13.6Z2
[1n2
f
− 1n2
i
]eV
Z = 1 for hydrogen
14Monday, April 2, 2007
Energy levels of the hydrogen atom – Bohr
Ene
rgy #
Ene
rgy #
Origin of the lines of the hydrogen atom spectrum
En =−13.6n2
eV
Ground state
Free electron
(Z = 1)
15Monday, April 2, 2007
Prob. 30.-/7: It is possible to use electromagnetic radiation to ionize atoms. To do so, the atoms must absorb the radiation, the photons of which must have enough energy to remove an electron from an atom.
What is the longest radiation wavelength that can be used to ionize the ground state of the hydrogen atom?
Ene
rgy #
Free electron
The least energy to remove a ground state electron is E& – E1.
Ground state,n = 1
16Monday, April 2, 2007
Ene
rgy #
Ground state
Free electron
First excited state, n = 2
Prob. 30.7/8: The electron in a hydrogen atom is in the first excited state, when the electron acquires an additional 2.86 eV of energy.
What is the quantum number, n, of the state into which the electron moves?
17Monday, April 2, 2007
Prob. 30.12/14: A hydrogen atom is in the ground state. It absorbs energy and makes a transition to the n = 3 excited state. The atom returns to the ground state emitting two photons. What are their wavelengths?
18Monday, April 2, 2007
This does NOT show the path of the
electron, but indicates there is a wave moving around the orbit –!take
with a pinch of salt!
Justification of Bohr’s condition on the angular momentum
Ln = mvr =nh
2!n = 1, 2, 3...
Substitute for the de Broglie wavelength of the electron:
mv= p=h
!
So, mvr =hr
!=nh
2"
" The circumference of the orbit is an integer number of wavelengths.
Condition for a “standing wave” that reinforces itself.
n = 4
Therefore, 2!r = n"
19Monday, April 2, 2007
Bohr’s theory applies to hydrogen-like atoms and ions – that have
just one electron in orbit around the nucleus. Examples:H atom, " Z = 1, 1 electron present
He+ ion, " Z = 2, 1 electron removedLi2+ ion, " Z = 3, 2 electrons removedBe3+ ion," Z = 4, 3 electrons removed, etc
Prob. 30.16: Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of lines in the lithium spectrum, the shortest wavelength if 40.5 nm. For the same series of lines in the beryllium spectrum, what is the shortest wavelength?
Both are hydrogen-like atoms, with all electrons removed but one, so Bohr’s formula applies for energies of stationary states:
" En = –13.6 Z2/n2 eV
20Monday, April 2, 2007
Spectrum of x-rays
• Sharp lines characteristic of the chemical " element of the target.• A smooth background (“Bremsstrahlung”)• A cutoff wavelength, $0, below which there are " no x-rays
An x-ray tube
Electrons: KE = eV
+
–Vacuum
21Monday, April 2, 2007
Spectrum of x-rays• Bremsstrahlung: due to the " electrons being stopped " rapidly on hitting the target – " accelerated charges emit " electromagnetic radiation.
• The shortest wavelength x-rays " are produced when all of the " kinetic energy of the electrons " is converted into x-rays:
• Characteristic x-rays: from electron transitions in the target atom. " Wavelengths depend on the target.
“L-shell”
! L!: (n = 3) " (n = 2)" L
%: (n = 4) " (n = 2) (Balmer), etc
“K-shell”
" K!: (n = 2) " (n = 1)" K
%: (n = 3) " (n = 1) (Lyman)
" KE = eV = hf = hc/$0
#o = 0.03 nmE = 41.4 keVV = 41.4 kV
22Monday, April 2, 2007
x-ray spectra at different voltages
• The characteristic x-ray peaks " do not move – they are " properties of the target atoms.
• The cutoff wavelength, $0,
& decreases as the accelerating " voltage increases:
& & $0 ∝ 1/V
• The Bremsstrahlung " component also shifts as the " voltage is changed.
eV =hc
!0
23Monday, April 2, 2007
n = 2
n = 1
K!
K shell
L shell
Wavelength of K! x-rays
Found that some elements were in the wrong place in the periodic table. Identified other previously unknown elements.
EK! =hc
"= 13.6(Z−1)2
(1
12− 1
22
)eV
Moseley found a quite accurate approximate result for K! x-rays for
atoms with one electron removed from the K shell:
30.34
The same as the Bohr formula, only with Z replaced by Z – 1.(there is a second electron in the K shell that partly shields the charge of the nucleus)
24Monday, April 2, 2007
Prob. 30.46/34: The atomic number of lead is Z = 82.
According to Moseley and the Bohr model, what is the energy of a K"
x-ray photon?
Prob. 30.36/48: What is the minimum potential difference that must be applied to an x-ray tube to knock a K-shell electron completely out of an atom of copper (Z = 29) target?
25Monday, April 2, 2007
A conventional x-ray picture, colour-enhanced
A picture of the shadows cast by bones and tissue.
The higher the accelerating voltage in the x-ray tube, the shorter the average wavelength of the x-rays and the more penetrating they are.
There is a trade-off between penetrating power of the x-rays and contrast of the image.
Lung tumour
26Monday, April 2, 2007
CT scans (Computerized Tomography) fire x-rays into the subject and creates x-ray shadow pictures from many directions. Computer analysis is able to reconstruct three-dimensional images from the many pictures. X-ray detectors are used in place of film.
In CAT scans (Computerized Axial Tomography) a narrow fan of x-rays is used to image a slice through the body.
27Monday, April 2, 2007
CAT Scan
(movable)
28Monday, April 2, 2007
http://en.wikipedia.org/wiki/Image:Venesreconstruction.JPG
Brain vessels reconstructed in 3D
29Monday, April 2, 2007
The LaserLight Amplification by
Stimulated Emission of Radiation
Spontaneous emission – the electron in an excited state falls back to a lower state at a random time, emitting a photon of energy E
i – E
f in a random direction.
E = Ei – E
f
E = Ei – E
f
E = Ei – E
f
Stimulated emission – a photon of energy E
i – E
f induces the
emission of a photon of the same energy, resulting in two photons of the same energy travelling in phase in the same direction
" amplification of the light.
30Monday, April 2, 2007
The Laser
For light amplification to occur:
• There must be a large number of atoms in the excited state – a " “population inversion”.
Metastable state (~10-3 s)
• The excited state must be long-lived – a “metastable state”.
Then atoms in the metastable state stay there long enough for a photon of the right energy to come along and cause stimulated emission.
The result is a build up of photons of the same wavelength that are in phase – coherent light.
E = Ei – Ef
31Monday, April 2, 2007
The Laser
Laser light is:
• Highly coherent. As the light waves are in phase, the combined " amplitude is large, and the intensity much greater than a source of " incoherent light, such as a lamp, that may emit the same number of " photons per second.
• Very sharply defined in wavelength, whereas an incandescent lamp " emits blackbody radiation, which is over a wide wavelength range, " including the invisible infra-red.
• Emitted in a narrow beam with divergence limited by diffraction.
32Monday, April 2, 2007
Helium-Neon Laser (eg checkout scanner)
Low pressure He (15%) - Ne (85%)
in a glass tube
A high voltage discharge in the gas produces a population inversion.
Parallel mirrors at the ends reflect light back and forth. Light angled too far from the axis of the laser escapes through the sides. Only light travelling parallel to the axis is amplified.
33Monday, April 2, 2007
Helium-Neon Laser
A high voltage discharge excites helium atoms to the 20.61 eV metastable state.
He and Ne atoms collide, the excitation is transferred to the Ne metastable level.
The energy of the photons is 20.66 – 18.70 eV = hc/$
$ = 633 nm, red.
Population inversion}
The population inversion is between the 20.66 and 18.70 eV levels of Ne.
34Monday, April 2, 2007
LasersThere are now many types of laser in power ranges from mW to MW, continuous beam and pulsed:
He-Ne, ruby, argon-ion, CO2, gallium arsenide solid state, chemical dye...
Uses:
• playback/recording of CDs, DVDs• checkout scanners• welding metal parts• accurate distance measurement• telecommunications• study of molecular structure• medicine" – selective removal of tissue, example, shaping the cornea of the eye" – removal of “port wine stain” birth marks" – destroying tumours with light-activated drugs" – reattaching detached retinas
35Monday, April 2, 2007
HolographyA laser beam is split into two parts. The first, a reference beam, travels directly to the film. The second reflects from the object and recombines with the reference beam at the film, generating interference fringes there.
Hologram (interference
fringes, not an image)
36Monday, April 2, 2007
l0
lm
Holography – the interference fringes
rm =√l2m− l2
0=
√(l0+m!)2− l2
0
rm =√m!(m!+2l0)!
√2ml0! "
√m
" the fringes are close together for large m
For constructive interference at P, the path difference between the reference beam and the beam that scatters from A should be an integer multiple of the wavelength of the light.
P
For the bright fringe of order m: lm – l0 = m$
[∆rm ∝ ∆m
2√
m
]
37Monday, April 2, 2007
Viewing the image by shining laser light on
the hologram
for first-order maximum
(the brightest)sin!=
"
d
d = spacing of the fringes
The fringes act as slits of varying spacingd
38Monday, April 2, 2007
Summary of Chapter 30
• The positive charge of the the atom is concentrated in a nucleus that is " ~10-15 - 10-14 m in radius.
• The angular momentum of the electron in an atom is quantized, " leading to quantized energy levels in hydrogen-like atoms:"
" " En = –13.6 Z2/n2 eV
• Photons are emitted and absorbed by atoms at the same wavelength" ! identification of elements in the “atmosphere” of stars, discovery of " " helium.
• The energies of K' x-rays can be calculated by replacing Z by Z–1 " ! identification of some chemical elements for the first time.
• Stimulated emission, the laser.
39Monday, April 2, 2007